InterviewSolution
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InterviewSolution
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If `f: R to R, g , R to R` be two funcitons, and `h(x) = 2 "min" {f(x) - g(x),0}` then `h(x)=`A. `f(x) + g(x) -|g(x)-f(x)|`B. `f(x) + g(x) +|g(x) - f(x)|`C. `f(x)-g(x)+|g(x) +|g(x) - f(x)|`D. `f(x) -g(x) -|g(x) - f(x)|` |
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Answer» Correct Answer - D Following cases arise. CASE-1 When `f(x) ge g(x)` In this case, we have `f(x)-g(x) ge 0` `therefore " ""min" {f(x) - g(x), 0 } = 0` `rArr h(x) = 0` `rArr h(x) = {f(x) - g(x)}-{f(x) -g(x)}` `rArr h(x) = {f (x) - g(x)} - |f (x) - g(x)|` `[{:(,because f(x) - g(x) ge 0),(,therefore|f(x) - g(x)|=f(x) - g(x)):}]` CASE II When `f(x) lt g(x)` In this case, we have `f(x) - g(x) lt 0` `therefore "min" {f(x) - g(x),0}=f(x) - g(x)` `rArr h(x) = 2 min {f(x) - g(x),0}=2 {f(x) - g(x)}` `rArr h (x) = {f(x) - g(x)} + {f(x) - g(x)}` `rArr h(x) = {f(x) - g(x)}-| f(x) - g(x)|`[{:(, therefore f(x) -g(x) lt 0),(,therefore |f(x) - g(x)|),(,=-{f(x) -g(x)}):}]` Hence, option (d) is correct . |
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